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3 years ago

8, 20, 50, ... Find the 10th term.

Mathematics

1 answer:

Murljashka [212]3 years ago

3 0

Answer:

$a_{10}=\frac{5^{10}}{2^7} -1$

Step-by-step explanation:

$a_{n}=a_1 . r^{n}-1$

Here

$r=\frac{5}{2}$

$a_1 =8$

For n=10 (tenth Term)

$a_{10}=8(\frac{5}{2})^{10} -1$

$a_{10}=\frac{5^{10}}{2^7} -1$

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Polar and cartesian equation<h2>Initial explanation</h2>

Let's analyze the relation between r and x and y:

We have that between the indicated value of r (of the polar coordinates) and x and y (of the cartesian coordinates) there is a relation because they form a triangle. If r changes, then the value of x and y will change.

<h2>STEP 1: given equation</h2>

Using the given equation

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we have that

$\begin{gathered} r=2secØ \\ \downarrow \\ \frac{r}{2}=secØ \end{gathered}$ <h2>STEP 2: secØ equation</h2>

Observing the image of the initial explanation we have a right triangle, we know that the equation of

secØ for any right triangle is given by:

$\sec Ø=\frac{\text{hypotenuse}}{\text{adjacent side}}$

In this case,

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then,

$\begin{gathered} \sec Ø=\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{r}{x} \\ \sec Ø=\frac{r}{x} \end{gathered}$ <h2>STEP 3: comparison between given equation and secØ equation</h2>

Then, we have that: